Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 20100;
int n, a[MAXN], b[MAXN];
vector<pair<int, int> > qa, qb;
void add(int* a, int i, int j, vector<pair<int, int> >& q) {
  a[i] ^= a[j];
  q.emplace_back(i, j);
}
void work(int* a, vector<pair<int, int> >& q) {
  for (int i = 30, j = 1; i >= 0; --i) {
    int k = j;
    while (k <= n && !(a[k] & (1 << i))) ++k;
    if (k > n) continue;
    if (k != j) add(a, j, k, q), add(a, k, j, q), add(a, j, k, q);
    for (int t = 1; t <= n; ++t)
      if ((a[t] & (1 << i)) && t != j) add(a, t, j, q);
    ++j;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= n; ++i) scanf("%d", b + i);
  work(a, qa);
  work(b, qb);
  for (int i = 1; i <= n && a[i]; ++i) {
    int k = 30;
    while (k >= 0 && !((1 << k) & a[i])) --k;
    for (int j = 1; j <= 30 && j <= n; ++j)
      if ((a[j] ^ b[j]) & (1 << k)) add(a, j, i, qa);
  }
  for (int i = 1; i <= n; ++i)
    if (a[i] != b[i]) {
      puts("-1");
      return 0;
    }
  printf("%d\n", qa.size() + qb.size());
  for (auto p : qa) printf("%d %d\n", p.first, p.second);
  reverse(qb.begin(), qb.end());
  for (auto p : qb) printf("%d %d\n", p.first, p.second);
}